If we say that an effectively generated first order theory $\sf T$ extends $\sf ZF$, such that every countable model of $\sf T$ doesn't have a class forcing extension that is pointwise definable. Would that just mean that $\sf T$ negates Choice? Or it does impart $\sf T$ proving some large cardinal property?
-
1$\begingroup$ When you say "forcing extension", does this include class forcing? That is what is needed for the arguments of my paper with Linetsky and Reitz. $\endgroup$– Joel David HamkinsCommented Jan 3, 2023 at 17:36
-
$\begingroup$ @JoelDavidHamkins, yes, of course. I’ve edited it. Thanks! $\endgroup$– Zuhair Al-JoharCommented Jan 3, 2023 at 19:03
1 Answer
If $T$ is a theory which proves "there is no extension of the model to a model of $\sf ZFC$ without adding ordinals", then there is no extension of models of $T$ by a class forcing to a pointwise definable model, since pointwise definable models must satisfy $\sf ZFC$.
The obvious example is Gitik's model, but we also have the Morris model where no large cardinals are involved, and for every $\alpha$ there is a set $A_\alpha$ which is the countable union of countable sets and $\mathcal P(A_\alpha)$ surjects onto $\omega_\alpha$. If we extended the Morris model to a model of $\sf ZFC$, then all the $A_\alpha$ became countable and all their power sets became the same size and therefore proper classes.
-
$\begingroup$ In the first line "there is no extension of the model" of what theory??? $\endgroup$ Commented Jan 4, 2023 at 9:33
-
-
$\begingroup$ You mean $T$ is an effectively generated FOL that extends ZFC and that proves that "there is no extension of a model of $T$ to a model of ZFC without adding ordinals". Isn't that inconsistent? Any model of $T$ extends to itself. $\endgroup$ Commented Jan 4, 2023 at 10:39
-
$\begingroup$ Your question is about theories extending ZF. I don't know how you conclude from "$T$ proves that there is no extension of the universe to a universe of ZFC without adding ordinals" to $T$ extends ZFC. $\endgroup$– Asaf Karagila ♦Commented Jan 4, 2023 at 10:59
-
1$\begingroup$ It implies there is a class generic extension satisfying AC, so in a sense, yes. $\endgroup$– Asaf Karagila ♦Commented Jan 4, 2023 at 11:18